如何使用规范化组仅当用户等于已验证用户时才获取用户电子邮件

Question

我用api- User创建了一个经典的平台类，我希望我的GET /users/{id}端点具有不同的行为，这取决于经过身份验证的用户是否等于{id}用户。

如果我是同一个用户(显示电子邮件)

{
  "@context": "/contexts/User",
  "@id": "/users/1",
  "@type": "User",
  "email": "peter@test.com",
  "username": "peter"
}

如果不是(隐藏电子邮件)

{
  "@context": "/contexts/User",
  "@id": "/users/1",
  "@type": "User",
  "username": "peter"
}

我尝试在电子邮件字段中添加自定义组，就像我可以在此处阅读https://api-platform.com/docs/core/serialization/#changing-the-serialization-context-dynamically一样

但是我不知道如何在SerializerContextBuilderInterface中检索user对象的信息

/**
 * @ORM\Column(type="string", length=180, unique=true)
 * @Groups({"owner:read", "user:write"})
 * @Assert\NotBlank()
 * @Assert\Email()
 */
private $email;

Answer 1

在实现SerializerContextBuilderInterface的类中，可以通过将TokenStorageInterface $tokenStorage传递给构造函数来检索当前用户。和对象请求(序列化所需的)，可以使用$user：$this->userRepository->find($request->get('id'));获取，示例如下：

public function createFromRequest(Request $request, bool $normalization, array $extractedAttributes = null) : array
{
    $context = $this->decorated->createFromRequest($request, $normalization, $extractedAttributes);

    $route = $request->get('_route');

    if ($route === 'api_users_get_item') {
        $currentUser = $this->token->getToken()->getUser();
        $userToSerialize = $this->userRepository->find($request->get('id'));
        if ($currentUser->getId() === $userToSerialize->getId()) {
            $context['groups'][] = 'owner:read';
        }
    }

    return $context;
}